Test Index

Circle

Section: Mathematics

© examsnet.com

Question : 13 of 89

Marks: +1, -0

If the equation of the polar of the point

(\alpha,-1)

with respect to the circle

x^2+y^2-4x-6y-12=0

y=\beta

4(\alpha+\beta)=

[AP EAPCET 23 May 2025 S1]

- 5
7
- 6
0

Solution:

Equation of polar is

T=0

\;\;\text{ i.e., }\; \alpha x+y(-1)-4 \cdot \frac{1}{2}(x+\alpha)

\;\;\; -6 \cdot \frac{1}{2}(y+(-1))-12=0

\; \Rightarrow \;\; \alpha x-y-2x-2\alpha-3y+3-12=0

\; \Rightarrow \;\; (\alpha-2x-4y-2\alpha-9=0 \;\;\; \cdot\cdot\cdot\cdot\cdot\cdot\cdot(i)

But equation of polar is

y=\beta

given

So, coefficient of

x

must be zero.

i.e.

\alpha-2=0 \Rightarrow \alpha=2

Thus,

0-4y-2 \times (2)-9=0(

from Eq.

(i))\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(ii)

\; \Rightarrow \;\; -4y=13

\; \Rightarrow \;\; y= \; \frac{-13}{4} = \beta ( \; \text{ given } \; )

\; \therefore 4(\alpha+\beta)=4 \left(2- \frac{13}{4}\right) = -5

© examsnet.com

Go to Question:

Prev Question

Next Question